maybe it helps to put $y = x-2$, so that $|y| \leq 1$.

Okay, I am back home from school.. problem doesn't look that hard, which means I am doing something wrong.. someone please check my soln? EDIT: The last part was bad, but I think its fixed now. Use above trick, substituting y = x+ 2. Now the coefficents of y are independent, so we can consider the equivalent problem of minimizing G, the absolute value of a general monic cubic $y^3 + ay^2 + by + c$ on the domain $|y| \leq 1$, over all reals a,b,c. Write it in the form: $ay^2 + c + (y^3 + by)$. It is easy to check that $(a,b,c) = (0,\frac{-3}{4},0)$ gives us a maximum of $1/4$, because $y^3 - (3/4)y = 1/4 \Leftrightarrow (y-1)(y+ 1/2)^2$. Suppose [for a contradiction] that (a,b,c) is a tuple that gives a lower maximum. By differentiation, it is clear that the extremal values of $y^3 + by$ occur at $y = \pm \sqrt{-b/3}$ (if b<0) and $y = \pm 1$, (in the domain $|y| \leq 1$). Now, $ay^2 + c \leq 0$ whereever maximums of $y^3 + by$ occur, since if it was $> 0$, we would have the maximum larger. Also, $ay^2 + c \geq 0$ whereever minimums of $y^3 + by$ occur. By symmetry, both in the places where extremal values of $y^3 + by$ occur, as well as the graph of $ay^2 + c$, we have to conclude $ay^2 + c = 0$ at the maximums of $y^3 + by$ (**). -- If b < -3/4, then y = 1 gives us a maximum > 1/4, and y = -1 gives us a minimum < 1/4. Since $ay^2 + c$ is an even function, if it is nonzero, then for some value of y in $\{-1, 1\}$, the signs of $ay^2 + c$ and $y^3 + by$ must be the same, and the magnitude of the entire cubic exceeds 1/4, if $ay^2 + c$ doesn't have roots at $y = \pm 1$. Now suppose $ay^2 + c$ has roots at $y = \pm 1$. Clearly, $y = \sqrt{-b/3}$ is a local maximum of $y^3 + by$, and by the supposition b < -3/4, we have infact that it is a real number in [0, 1/2]. Now consider $\sqrt{-b/3}^3 + -ab/3 + b\sqrt{-b/3} + c$. By "$ay^2 + c$ has roots at $y = \pm 1$", we have $-a = c$; by considering the cubic at $y=0$, we have $|a| < 1/4$. So we have $\sqrt{-b/3}(4b/3) + c(b/3 + 1) < (1/2)(-1) + c(3/4) = 3c/4 - 1/2$. Now $|3c/4| < 3/16$, whereby $3c/4 - 1/2 < -5/16$. So the general cubic has a magnitude above $1/4$ at that point, and it is inadmissible. Similarily, consider b > -3/4 at the point $y = \sqrt{-b/3}$. Similarily, we must have $ay^2 + c$ with roots at $y = \pm \sqrt{-b/3}$. This implies ab = 3c. Consider $y = 1$ in the general cubic. We have $1 + a + b + c > 1/4 + a + c$. Now $3c = ab > -3a/4$, whereby $c > -a/4$. So $1/4 + a + c > 1/4 + 3a/4$. Obviously, due to the symmetry (oddness) of $y^3 + by$, it doesn't matter whether a was positive or negative. So the magnitude of the entire cubic is clearly $> 1/4$, and so it is inadmissible. So $b = -3/4$. By (**), $ay^2 + c$ is zero at 4 places, thus it must be the zero function. So we have $(a,b,c) = (0, -3/4, 0)$, with a maximum of 1/4. This is the same maximum as in the original problem, so we are done.

I think I solved a much more general problem of Alexandru Lupas posted on the forum and which asked for the minimal value of a monic polynomial on a compact interval $[a,b]$. Anyway, it is a direct consequence of Chebyshev's theorem for polynomials on $[-1,1]$.

Substituting $y = x+ 2$ Let $f(y)=y^3+Ay^2+By+C (-1\leq y\leq 1)$ $f(-1)=-1+A-B+C$ $-2f(-\frac{1}{2})=1/4-A/2+B-2C$ $2f(\frac{1}{2})=1/4+A/2+B+2C$ $-f(1)=-1-A-B-C$ then $6F \geq |f(-1)-2f(-\frac{1}{2})+2f(\frac{1}{2})-f(1)| = \frac{3}{2}$ thus $F \geq 1/4$ then it is easy to prove $|f(y)|=|x^3-\frac{3}{4}x| \geq \frac{1}{4}$

Chebyshev's theorem Let $f \in \mathbb{R}[x]$ be a monic polynomial of degree $n$. Then $$\max_{x\in [-1,1]}|f(x)| \ge \frac{1}{2^{n-1}}$$ Original problem: Denote $y=x-2$. $(y\in [-1,1]) \implies F=\max_{y \in [-1,1]} |y^3 +Ax^2 +Bx+C|$ By Chebyshev's theorem, $$F\ge \frac{1}{2^{3-1}}=\frac{1}{4}$$Answer: $\frac{1}{4}$

Equality holds $f(x)=\frac{1}{4}T_{3}(x)$. $T_{3}(x)$ is Chebyshev polynomials $(T_{3}(x)=4x^3-3x) (\because cos3x=4cos^3x-3cosx)$